Question

If $\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+c=0$, then $\frac{\alpha }{\alpha \beta +b}+\frac{\beta }{a\alpha +b}$ is equal to

• A. $\frac{2}{a}$
• B. $\frac{2}{b}$
• C. $\frac{2}{c}$
• D. $-\frac{2}{a}$

Answer 1

The correct option is D $-\frac{2}{a}$

$\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$ and ${\alpha }^2+{\beta }^2=\frac{(b^2-2ac)}{a^2}$

Now, $\frac{\alpha }{a\beta +b}+\frac{\beta }{a\alpha +b}=\frac{\alpha (a\alpha +b)+\beta (a\beta +b)}{(a\beta +b)(a\alpha +b)}$

$=\frac{\alpha ({\alpha }^2+{\beta }^2)+b(\alpha +\beta )}{\alpha \beta a^2+b(\alpha +\beta )+b^2}=\frac{a\frac{(b^2-2ac)}{a^2}+b[-\frac{b}{a}]}{\left[\frac{c}{a}\right]a^2+ab[-\frac{b}{a}]+b^2}$

$=\frac{(b^2-2ac-b^2)}{a^{2}c-a{b}^2+a{b}^2}=\frac{-2ac}{a^{2}c}=-\frac{2}{a}$